Codeforces Round 935 (Div. 3)A~E

A. Setting up Camp

题目分析:

 有三种人，内向、外向、综合，内向必须独自一个帐篷，外向必须3个人一个帐篷，综合介于1~3人一个帐篷，我们发现非法情况只会存在外向的人凑不成3个人一个帐篷的情况，因外向不够可以向综合人借，故将二者合并并判断:

#include<bits/stdc++.h>
#pragma GCC optimize("Ofast")
#define INF 1e18
#define IOS ios::sync_with_stdio(false);cin.tie(0);
#define int long long
#define pb push_back
#define vct vector
#define checkbit __builtin_popcount
#define gcd __gcd
#define use int T;cin>>T;while(T--)
#define LEN length()
#define all(a) a.begin()+1,a.end()
template<class T> bool mmax(T &u, T v) { return u < v ? (u = v, 1) : 0; }
template<class T> bool mmin(T &u, T v) { return u > v ? (u = v, 1) : 0; }
#define lowbit(x) (x&(-x))
#define yes cout<<"YES"<<'\n'
#define no cout<<"NO"<<'\n' 
#define safe_map unordered_map<int, int, custom_hash>
using namespace std;
typedef pair<int,int>pii;
const int N =2e5+7;
struct custom_hash {
size_t operator()(uint64_t x) const {
static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
x ^= FIXED_RANDOM;
return x ^ (x >> 16);
}
};
signed main()
{IOSuse{int a,b,c;cin>>a>>b>>c;int x=b%3;if(x+c<3&&x!=0)cout<<"-1"<<endl;// 非法情况else cout<<a+(b+c)/3+((b+c)%3?1:0)<<endl;//else}
return 0;
}

B. Fireworks

题目分析:

A类烟花每a分钟发射一次，B类烟花每b分钟发射一次，它们都能持续m分钟,那么我们直接用m分别求出这两类烟花都能放几个加和即可，注意， 我们可以选择烟花发射时间包含二者最小公倍数所以最终结果+2

#include<bits/stdc++.h>
#pragma GCC optimize("Ofast")
#define INF 1e18
#define IOS ios::sync_with_stdio(false);cin.tie(0);
#define int long long
#define pb push_back
#define vct vector
#define checkbit __builtin_popcount
#define gcd __gcd
#define use int T;cin>>T;while(T--)
#define LEN length()
#define all(a) a.begin()+1,a.end()
template<class T> bool mmax(T &u, T v) { return u < v ? (u = v, 1) : 0; }
template<class T> bool mmin(T &u, T v) { return u > v ? (u = v, 1) : 0; }
#define lowbit(x) (x&(-x))
#define yes cout<<"YES"<<'\n'
#define no cout<<"NO"<<'\n' 
#define safe_map unordered_map<int, int, custom_hash>
using namespace std;
typedef pair<int,int>pii;
const int N =2e5+7;
struct custom_hash {
size_t operator()(uint64_t x) const {
static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
x ^= FIXED_RANDOM;
return x ^ (x >> 16);
}
};
signed main()
{IOSuse{int a,b,m;cin>>a>>b>>m;cout<<m/a+m/b+2<<endl;}return 0;
}

C. Left and Right Houses 

题目分析:

给定字符串s,要求将字符串分隔，并且确保满足题目条件，遍历即可

#include<bits/stdc++.h>
#pragma GCC optimize("Ofast")
#define INF 1e18
#define IOS ios::sync_with_stdio(false);cin.tie(0);
#define int long long
#define pb push_back
#define vct vector
#define checkbit __builtin_popcount
#define gcd __gcd
#define use int T;cin>>T;while(T--)
#define LEN length()
#define all(a) a.begin()+1,a.end()
template<class T> bool mmax(T &u, T v) { return u < v ? (u = v, 1) : 0; }
template<class T> bool mmin(T &u, T v) { return u > v ? (u = v, 1) : 0; }
#define lowbit(x) (x&(-x))
#define yes cout<<"YES"<<'\n'
#define no cout<<"NO"<<'\n' 
#define safe_map unordered_map<int, int, custom_hash>
using namespace std;
typedef pair<int,int>pii;
const int N =2e5+7;
struct custom_hash {
size_t operator()(uint64_t x) const {
static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
x ^= FIXED_RANDOM;
return x ^ (x >> 16);
}
};
char s[3000005];
int s1[300005];
signed main()
{
use{int n;cin>>n;scanf("%s",s+1);for(int i=1;i<=n;i++)s1[i]=s1[i-1]+(s[i]-48);int ans=-1;for(int i=0;i<=n;i++)if(s1[i]<=i-s1[i] && s1[n]-s1[i]>=(n-i)-(s1[n]-s1[i])){if(abs(n-i-i)<abs(n-ans-ans))ans=i;}cout<<ans<<endl;
}return 0;
}

D. Seraphim the Owl 

题目分析:

当主人公在i位置时，他想要换到j位置，首先需要花费 a[j] 元, 并且需要花费 

因为主人公需要至少排在第m位，所以在n+1~m-1之间我们可以选择最优值，也就是min(a[i],b[i]),

到m后，我们可以选择a[m]来停下，但是考虑到有可能到前面的位置花费会更小所以我们再向前模拟，更新最小值

#include<bits/stdc++.h>
#pragma GCC optimize("Ofast")
#define INF 1e18
#define IOS ios::sync_with_stdio(false);cin.tie(0);
#define int long long
#define pb push_back
#define vct vector
#define checkbit __builtin_popcount
#define gcd __gcd
#define use int T;cin>>T;while(T--)
#define LEN length()
#define all(a) a.begin()+1,a.end()
template<class T> bool mmax(T &u, T v) { return u < v ? (u = v, 1) : 0; }
template<class T> bool mmin(T &u, T v) { return u > v ? (u = v, 1) : 0; }
#define lowbit(x) (x&(-x))
#define yes cout<<"YES"<<'\n'
#define no cout<<"NO"<<'\n' 
#define safe_map unordered_map<int, int, custom_hash>
using namespace std;
typedef pair<int,int>pii;
const int N =2e5+7;
struct custom_hash {
size_t operator()(uint64_t x) const {
static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
x ^= FIXED_RANDOM;
return x ^ (x >> 16);
}
};
signed main()
{IOSuse{int n,m;cin>>n>>m;vct<int>a(n+1),b(n+1);for(int i=1;i<=n;i++)cin>>a[i];for(int i=1;i<=n;i++)cin>>b[i];int ans=0;for(int i=n;i>m;i--){ans+=min(a[i],b[i]);}int mins=a[m];int x=0;for(int i=m;i>1;i--){x+=b[i];mmin(mins,x+a[i-1]);}cout<<ans+mins<<'\n';}return 0;
}

E. Binary Search

 

题目分析:

二分但没完全二分，数组的取值为1~n,且可能不单调，题目给出的方法适用于单调递增，故，如果起初数组为单调递增的话操作数为0，否则，我们按照题意进行二分模拟，最终二分出来的坐标l 与x起初的位置换一次即可；

证明可行性,因在原数组模拟的情况下，在过程中l满足小于等于x的性质，x也同样满足小于等于x的性质，故可等价替换，故仅需要一次操作即可二分得到x

#include<bits/stdc++.h>
#pragma GCC optimize("Ofast")
#define INF 1e18
#define IOS ios::sync_with_stdio(false);cin.tie(0);
#define int long long
#define pb push_back
#define vct vector
#define checkbit __builtin_popcount
#define gcd __gcd
#define use int T;cin>>T;while(T--)
#define LEN length()
#define all(a) a.begin()+1,a.end()
template<class T> bool mmax(T &u, T v) { return u < v ? (u = v, 1) : 0; }
template<class T> bool mmin(T &u, T v) { return u > v ? (u = v, 1) : 0; }
#define lowbit(x) (x&(-x))
#define yes cout<<"YES"<<'\n'
#define no cout<<"NO"<<'\n' 
#define safe_map unordered_map<int, int, custom_hash>
using namespace std;
typedef pair<int,int>pii;
const int N =2e5+7;
struct custom_hash {
size_t operator()(uint64_t x) const {
static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
x ^= FIXED_RANDOM;
return x ^ (x >> 16);
}
};
signed main()
{IOSuse{int n,x;cin>>n>>x;vct<int>a(n+1);int cnt=0;safe_map mp;for(int i=1;i<=n;i++){cin>>a[i];mp[a[i]]=i;}if(is_sorted(all(a)))cout<<"0"<<endl;else {cout<<"1"<<endl;int l=1,r=n+1;while(l<r-1){int mid =(l+r)>>1;if(a[mid]>x)r=mid;else l=mid;}cout<<l<<" "<<mp[x]<<endl;}}return 0;
}

 总结:A~E题都可以用模拟or 思维解出.